$2^n + 1$ is prime $ implies n$ is a power of $2$












2














I was wondering why I can't do this this way by proof by contradiction of the contrapositive. So I want to prove

$$2^n + 1 quad text{is prime} implies n = 2^k quad text{for some} kin mathbb{Z}.$$

By contrapositive,

$$n neq 2^k quad text{for all} k in mathbb{Z} implies 2^n + 1 quad text{is composite.}$$



Then I want to prove by contradiction:

Suppose $n neq 2^k quad forall kin mathbb{Z}$ and $2^n + 1$ is prime.

Why can't I give one counterexample to prove that this is false (contradiction)?

Since if $n:= 3$ then $2^3 + 1 = 9$ which is composite, hence, $2^n + 1$ being prime (hypothesis) cannot be true, so by proof by contradiction, $2^n +1$ must be composite?






elementary-number-theory proof-verification prime-numbers fermat-numbers







share|cite|improve this question




















  • 2




    Because you have to prove it for all $k in mathbb{Z}$.
    – Math Lover
    Aug 15 '17 at 4:18












  • Strictly speaking you also need $ngt 0$, because $2^0 + 1 = 2$ is prime (but $0$ is not a power of $2$).
    – hardmath
    Aug 15 '17 at 4:33






  • 1




    You want to "prove" that All Nordic women have blonde hair by exhibiting a Chinese lady with black hair?
    – Jyrki Lahtonen
    Aug 15 '17 at 4:45












  • @Math Lover I think it should be for all $kinmathbb N$.
    – Michael Rozenberg
    Aug 15 '17 at 4:53






  • 3




    Possible duplicate of Fermat primes relation to $2^n+1$
    – Gerry Myerson
    Aug 19 '17 at 22:26
















2














I was wondering why I can't do this this way by proof by contradiction of the contrapositive. So I want to prove

$$2^n + 1 quad text{is prime} implies n = 2^k quad text{for some} kin mathbb{Z}.$$

By contrapositive,

$$n neq 2^k quad text{for all} k in mathbb{Z} implies 2^n + 1 quad text{is composite.}$$



Then I want to prove by contradiction:

Suppose $n neq 2^k quad forall kin mathbb{Z}$ and $2^n + 1$ is prime.

Why can't I give one counterexample to prove that this is false (contradiction)?

Since if $n:= 3$ then $2^3 + 1 = 9$ which is composite, hence, $2^n + 1$ being prime (hypothesis) cannot be true, so by proof by contradiction, $2^n +1$ must be composite?






elementary-number-theory proof-verification prime-numbers fermat-numbers







share|cite|improve this question




















  • 2




    Because you have to prove it for all $k in mathbb{Z}$.
    – Math Lover
    Aug 15 '17 at 4:18












  • Strictly speaking you also need $ngt 0$, because $2^0 + 1 = 2$ is prime (but $0$ is not a power of $2$).
    – hardmath
    Aug 15 '17 at 4:33






  • 1




    You want to "prove" that All Nordic women have blonde hair by exhibiting a Chinese lady with black hair?
    – Jyrki Lahtonen
    Aug 15 '17 at 4:45












  • @Math Lover I think it should be for all $kinmathbb N$.
    – Michael Rozenberg
    Aug 15 '17 at 4:53






  • 3




    Possible duplicate of Fermat primes relation to $2^n+1$
    – Gerry Myerson
    Aug 19 '17 at 22:26














2












2








2







I was wondering why I can't do this this way by proof by contradiction of the contrapositive. So I want to prove

$$2^n + 1 quad text{is prime} implies n = 2^k quad text{for some} kin mathbb{Z}.$$

By contrapositive,

$$n neq 2^k quad text{for all} k in mathbb{Z} implies 2^n + 1 quad text{is composite.}$$



Then I want to prove by contradiction:

Suppose $n neq 2^k quad forall kin mathbb{Z}$ and $2^n + 1$ is prime.

Why can't I give one counterexample to prove that this is false (contradiction)?

Since if $n:= 3$ then $2^3 + 1 = 9$ which is composite, hence, $2^n + 1$ being prime (hypothesis) cannot be true, so by proof by contradiction, $2^n +1$ must be composite?






elementary-number-theory proof-verification prime-numbers fermat-numbers







share|cite|improve this question















I was wondering why I can't do this this way by proof by contradiction of the contrapositive. So I want to prove

$$2^n + 1 quad text{is prime} implies n = 2^k quad text{for some} kin mathbb{Z}.$$

By contrapositive,

$$n neq 2^k quad text{for all} k in mathbb{Z} implies 2^n + 1 quad text{is composite.}$$



Then I want to prove by contradiction:

Suppose $n neq 2^k quad forall kin mathbb{Z}$ and $2^n + 1$ is prime.

Why can't I give one counterexample to prove that this is false (contradiction)?

Since if $n:= 3$ then $2^3 + 1 = 9$ which is composite, hence, $2^n + 1$ being prime (hypothesis) cannot be true, so by proof by contradiction, $2^n +1$ must be composite?






elementary-number-theory proof-verification prime-numbers fermat-numbers




elementary-number-theory proof-verification prime-numbers fermat-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 21:52









greedoid

38.6k114797




38.6k114797










asked Aug 15 '17 at 4:15









OneGapLaterOneGapLater

297314




297314








  • 2




    Because you have to prove it for all $k in mathbb{Z}$.
    – Math Lover
    Aug 15 '17 at 4:18












  • Strictly speaking you also need $ngt 0$, because $2^0 + 1 = 2$ is prime (but $0$ is not a power of $2$).
    – hardmath
    Aug 15 '17 at 4:33






  • 1




    You want to "prove" that All Nordic women have blonde hair by exhibiting a Chinese lady with black hair?
    – Jyrki Lahtonen
    Aug 15 '17 at 4:45












  • @Math Lover I think it should be for all $kinmathbb N$.
    – Michael Rozenberg
    Aug 15 '17 at 4:53






  • 3




    Possible duplicate of Fermat primes relation to $2^n+1$
    – Gerry Myerson
    Aug 19 '17 at 22:26














  • 2




    Because you have to prove it for all $k in mathbb{Z}$.
    – Math Lover
    Aug 15 '17 at 4:18












  • Strictly speaking you also need $ngt 0$, because $2^0 + 1 = 2$ is prime (but $0$ is not a power of $2$).
    – hardmath
    Aug 15 '17 at 4:33






  • 1




    You want to "prove" that All Nordic women have blonde hair by exhibiting a Chinese lady with black hair?
    – Jyrki Lahtonen
    Aug 15 '17 at 4:45












  • @Math Lover I think it should be for all $kinmathbb N$.
    – Michael Rozenberg
    Aug 15 '17 at 4:53






  • 3




    Possible duplicate of Fermat primes relation to $2^n+1$
    – Gerry Myerson
    Aug 19 '17 at 22:26








2




2




Because you have to prove it for all $k in mathbb{Z}$.
– Math Lover
Aug 15 '17 at 4:18






Because you have to prove it for all $k in mathbb{Z}$.
– Math Lover
Aug 15 '17 at 4:18














Strictly speaking you also need $ngt 0$, because $2^0 + 1 = 2$ is prime (but $0$ is not a power of $2$).
– hardmath
Aug 15 '17 at 4:33




Strictly speaking you also need $ngt 0$, because $2^0 + 1 = 2$ is prime (but $0$ is not a power of $2$).
– hardmath
Aug 15 '17 at 4:33




1




1




You want to "prove" that All Nordic women have blonde hair by exhibiting a Chinese lady with black hair?
– Jyrki Lahtonen
Aug 15 '17 at 4:45






You want to "prove" that All Nordic women have blonde hair by exhibiting a Chinese lady with black hair?
– Jyrki Lahtonen
Aug 15 '17 at 4:45














@Math Lover I think it should be for all $kinmathbb N$.
– Michael Rozenberg
Aug 15 '17 at 4:53




@Math Lover I think it should be for all $kinmathbb N$.
– Michael Rozenberg
Aug 15 '17 at 4:53




3




3




Possible duplicate of Fermat primes relation to $2^n+1$
– Gerry Myerson
Aug 19 '17 at 22:26




Possible duplicate of Fermat primes relation to $2^n+1$
– Gerry Myerson
Aug 19 '17 at 22:26










2 Answers
2






active

oldest

votes


















4














The question isn't to show that there is one $n$ that is not a power of $2$ with $2^n+1$ composite, it is to show that for every $n$ that is not a power of $2$, $2^n+1$ is composite.






share|cite|improve this answer





















  • I see; would my contrapositive and then contradiction (or just contrapositive) path lead to anything?
    – OneGapLater
    Aug 15 '17 at 4:20












  • Yes, that is indeed the typical route to go about proving this statement.
    – Ziryerx
    Aug 15 '17 at 5:47



















3














Suppose there is a odd prime $p|n$. Then $n=pcdot k$ for some natural $k$. Now we have:
$$ 2^{pk}+1 =(2^k+1)(2^{k(p-1)}-2^{k(p-2)}+...-2^k+1)$$



Clearly both factors are > 1 and thus a contradction.






share|cite|improve this answer





















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    2 Answers
    2






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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

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    4














    The question isn't to show that there is one $n$ that is not a power of $2$ with $2^n+1$ composite, it is to show that for every $n$ that is not a power of $2$, $2^n+1$ is composite.






    share|cite|improve this answer





















    • I see; would my contrapositive and then contradiction (or just contrapositive) path lead to anything?
      – OneGapLater
      Aug 15 '17 at 4:20












    • Yes, that is indeed the typical route to go about proving this statement.
      – Ziryerx
      Aug 15 '17 at 5:47
















    4














    The question isn't to show that there is one $n$ that is not a power of $2$ with $2^n+1$ composite, it is to show that for every $n$ that is not a power of $2$, $2^n+1$ is composite.






    share|cite|improve this answer





















    • I see; would my contrapositive and then contradiction (or just contrapositive) path lead to anything?
      – OneGapLater
      Aug 15 '17 at 4:20












    • Yes, that is indeed the typical route to go about proving this statement.
      – Ziryerx
      Aug 15 '17 at 5:47














    4












    4








    4






    The question isn't to show that there is one $n$ that is not a power of $2$ with $2^n+1$ composite, it is to show that for every $n$ that is not a power of $2$, $2^n+1$ is composite.






    share|cite|improve this answer












    The question isn't to show that there is one $n$ that is not a power of $2$ with $2^n+1$ composite, it is to show that for every $n$ that is not a power of $2$, $2^n+1$ is composite.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Aug 15 '17 at 4:19









    Robert IsraelRobert Israel

    319k23208458




    319k23208458












    • I see; would my contrapositive and then contradiction (or just contrapositive) path lead to anything?
      – OneGapLater
      Aug 15 '17 at 4:20












    • Yes, that is indeed the typical route to go about proving this statement.
      – Ziryerx
      Aug 15 '17 at 5:47


















    • I see; would my contrapositive and then contradiction (or just contrapositive) path lead to anything?
      – OneGapLater
      Aug 15 '17 at 4:20












    • Yes, that is indeed the typical route to go about proving this statement.
      – Ziryerx
      Aug 15 '17 at 5:47
















    I see; would my contrapositive and then contradiction (or just contrapositive) path lead to anything?
    – OneGapLater
    Aug 15 '17 at 4:20






    I see; would my contrapositive and then contradiction (or just contrapositive) path lead to anything?
    – OneGapLater
    Aug 15 '17 at 4:20














    Yes, that is indeed the typical route to go about proving this statement.
    – Ziryerx
    Aug 15 '17 at 5:47




    Yes, that is indeed the typical route to go about proving this statement.
    – Ziryerx
    Aug 15 '17 at 5:47











    3














    Suppose there is a odd prime $p|n$. Then $n=pcdot k$ for some natural $k$. Now we have:
    $$ 2^{pk}+1 =(2^k+1)(2^{k(p-1)}-2^{k(p-2)}+...-2^k+1)$$



    Clearly both factors are > 1 and thus a contradction.






    share|cite|improve this answer


























      3














      Suppose there is a odd prime $p|n$. Then $n=pcdot k$ for some natural $k$. Now we have:
      $$ 2^{pk}+1 =(2^k+1)(2^{k(p-1)}-2^{k(p-2)}+...-2^k+1)$$



      Clearly both factors are > 1 and thus a contradction.






      share|cite|improve this answer
























        3












        3








        3






        Suppose there is a odd prime $p|n$. Then $n=pcdot k$ for some natural $k$. Now we have:
        $$ 2^{pk}+1 =(2^k+1)(2^{k(p-1)}-2^{k(p-2)}+...-2^k+1)$$



        Clearly both factors are > 1 and thus a contradction.






        share|cite|improve this answer












        Suppose there is a odd prime $p|n$. Then $n=pcdot k$ for some natural $k$. Now we have:
        $$ 2^{pk}+1 =(2^k+1)(2^{k(p-1)}-2^{k(p-2)}+...-2^k+1)$$



        Clearly both factors are > 1 and thus a contradction.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 15 '17 at 6:07









        greedoidgreedoid

        38.6k114797




        38.6k114797






























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